What is the time complexity of Handling Sum Queries After Update?

The optimal solution has a time complexity of O(q*log(n)).

What data structures are used to solve Handling Sum Queries After Update?

This problem uses Array, Segment Tree. Understanding these concepts is essential for solving this Hard problem efficiently.

What are the key insights for solving Handling Sum Queries After Update?

Insight 1: Using a naive approach of iterating through the range for each update (type 1 query) would lead to a time complexity of O(n*q), where n is the length of the arrays and q is the number of queries, potentially exceeding the time limit for larger inputs. Therefore, a Segment Tree with Lazy Propagation is crucial. Insight 2: The Segment Tree allows for efficient range updates (flips in this case) and range queries (sum of 1s). Lazy propagation ensures that updates are applied to the tree nodes only when necessary, optimizing the process.

Which companies ask LeetCode 2569. Handling Sum Queries After Update in interviews?

This problem is frequently asked at Trilogy.

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Handling Sum Queries After Update - LeetCode 2569 Solution

Q: How to solve Handling Sum Queries After Update?

The provided code uses a Segment Tree with Lazy Propagation to efficiently handle the queries. The segment tree stores the sum of 1's within a given range of `nums1`. Lazy propagation is used to defer the actual flipping of bits until necessary (i.e., when querying or updating a node's children). The `nums2` array's sum is also maintained, allowing O(1) time complexity for answering type 3 queries.

Handling Sum Queries After Update - Complete Solution Guide

Handling Sum Queries After Update is LeetCode problem 2569, a Hard level challenge. This complete guide provides step-by-step explanations, multiple solution approaches, and optimized code in python3, java, cpp, c.

Problem Statement

You are given two 0-indexed arrays nums1 and nums2 and a 2D array queries of queries. There are three types of queries: For a query of type 1, queries[i] = [1, l, r] . Flip the values from 0 to 1 and from 1 to 0 in nums1 from index l to index r . Both l and r are 0-indexed . For a query of type 2, queries[i] = [2, p, 0] . For every index 0 <= i < n , set nums2[i] = nums2[i] + nums1[i] * p . For a query of type 3, queries[i] = [3, 0, 0] . Find the sum of the elements in nums2 . Return an array co

Detailed Explanation

The problem presents two arrays, `nums1` (containing 0s and 1s) and `nums2` (containing non-negative integers). It also provides a list of queries. There are three types of queries: 1. **Type 1 (Flip):** Given a range `[l, r]`, flip the bits in `nums1` within that range (0 becomes 1, and 1 becomes 0). 2. **Type 2 (Update nums2):** Given a value `p`, update each element `nums2[i]` by adding `nums1[i] * p` to it. 3. **Type 3 (Sum nums2):** Calculate the sum of all elements in `nums2`. The goal is to process the queries and return an array containing the results of all Type 3 queries.

Solution Approach

The provided code uses a Segment Tree with Lazy Propagation to efficiently handle the queries. The segment tree stores the sum of 1's within a given range of `nums1`. Lazy propagation is used to defer the actual flipping of bits until necessary (i.e., when querying or updating a node's children). The `nums2` array's sum is also maintained, allowing O(1) time complexity for answering type 3 queries.

Step-by-Step Algorithm

Step 1: **Build the Segment Tree:** Create a segment tree from the `nums1` array. Each node in the tree represents a range of indices in `nums1` and stores the sum of 1s within that range.
Step 2: **Implement Lazy Propagation:** Maintain a `lazy` array to store pending updates (flips) for each node in the segment tree. A value of 1 in `lazy[i]` indicates that the corresponding node needs to be flipped.
Step 3: **Implement `push` function:** When visiting a node, push any pending updates (from `lazy`) to its children and update the node's value. Also, clear the lazy flag for the current node after propagating it.
Step 4: **Implement `update` function:** For Type 1 queries (flip range), use the segment tree's update function to flip the bits within the given range. This involves traversing the tree, applying lazy propagation where needed, and updating the tree nodes.
Step 5: **Process Type 2 queries:** For Type 2 queries (update nums2), multiply the sum of 1s in `nums1` (obtained from the root of the segment tree) by `p` and add it to the running sum of `nums2`.
Step 6: **Process Type 3 queries:** For Type 3 queries (sum nums2), simply append the current sum of `nums2` to the result array.
Step 7: **Return results:** Finally, return the array containing the results of all Type 3 queries.

Key Insights

Insight 1: Using a naive approach of iterating through the range for each update (type 1 query) would lead to a time complexity of O(n*q), where n is the length of the arrays and q is the number of queries, potentially exceeding the time limit for larger inputs. Therefore, a Segment Tree with Lazy Propagation is crucial.
Insight 2: The Segment Tree allows for efficient range updates (flips in this case) and range queries (sum of 1s). Lazy propagation ensures that updates are applied to the tree nodes only when necessary, optimizing the process.
Insight 3: The sum of nums2 can be maintained throughout the operations, updated each time a type 2 query happens.

Complexity Analysis

Time Complexity: O(q*log(n))

Space Complexity: O(n)

Topics

This problem involves: Array, Segment Tree.

Companies

Asked at: Trilogy.

Handling Sum Queries After Update | LeetCode 2569

Hard

Array Segment Tree

Time: O(q*log(n))Space: O(n)

Solution Code

Available in:

class Solution {
    private int[] tree;
    private int[] lazy;
    private int n;

    public long[] handleQuery(int[] nums1, int[] nums2, int[][] queries) {
        n = nums1.length;
        tree = new int[4 * n];
        lazy = new int[4 * n];

        build(1, 0, n - 1, nums1);

        long sum2 = 0;
        for (int num : nums2) {
            sum2 += num;
        }

        List<Long> ansList = new ArrayList<>();

        for (int[] query : queries) {
            int q_type = query[0];
            if (q_type == 1) {
                int l = query[1];
                int r = query[2];
                update(1, 0, n - 1, l, r);
            } else if (q_type == 2) {
                int p = query[1];
                int count_ones = tree[1];
                sum2 += (long)p * count_ones;
            } else if (q_type == 3) {
                ansList.add(sum2);
            }
        }

        long[] ans = new long[ansList.size()];
        for (int i = 0; i < ansList.size(); i++) {
            ans[i] = ansList.get(i);
        }

        return ans;
    }

    private void build(int node, int start, int end, int[] nums1) {
        if (start == end) {
            tree[node] = nums1[start];
            return;
        }
        int mid = (start + end) / 2;
        int left_node = 2 * node;
        int right_node = 2 * node + 1;
        build(left_node, start, mid, nums1);
        build(right_node, mid + 1, end, nums1);
        tree[node] = tree[left_node] + tree[right_node];
    }

    private void push(int node, int start, int end) {
        if (lazy[node] == 1) {
            int range_len = end - start + 1;
            tree[node] = range_len - tree[node];
            if (start != end) {
                int left_node = 2 * node;
                int right_node = 2 * node + 1;
                lazy[left_node] ^= 1;
                lazy[right_node] ^= 1;
            }
            lazy[node] = 0;
        }
    }

    private void update(int node, int start, int end, int l, int r) {
        push(node, start, end);

        if (start > r || end < l) {
            return;
        }

        if (l <= start && end <= r) {
            lazy[node] ^= 1;
            push(node, start, end);
            return;
        }

        int mid = (start + end) / 2;
        int left_node = 2 * node;
        int right_node = 2 * node + 1;
        update(left_node, start, mid, l, r);
        update(right_node, mid + 1, end, l, r);

        tree[node] = tree[left_node] + tree[right_node];
    }
}

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Problem Info

Description

Problem 2569: Handling Sum Queries After Update - Detailed explanation and solution approaches.

Topics

Array Segment Tree

Companies

Trilogy

Stats

Difficulty:Hard

Languages:4

Companies:1

Problem #:2569

Problem Explanation

💡

Think Before You Code

🤔 Ask Yourself

→What is the input? What output do I need?
→Can I solve a smaller version of this problem first?
→What patterns have I seen in similar problems?

✏️ Try This First

Before looking at the solution, try to solve it on paper with a simple example!

🚀 Try on LeetCode

Solution Approach

📋

Step-by-Step Solution

7 steps

1
Step 1: **Build the Segment Tree:** Create a segment tree from the `nums1` array. Each node in the tree represents a range of indices in `nums1` and stores the sum of 1s within that range.
2
Step 2: **Implement Lazy Propagation:** Maintain a `lazy` array to store pending updates (flips) for each node in the segment tree. A value of 1 in `lazy[i]` indicates that the corresponding node needs to be flipped.
3
Step 3: **Implement `push` function:** When visiting a node, push any pending updates (from `lazy`) to its children and update the node's value. Also, clear the lazy flag for the current node after propagating it.
4
Step 4: **Implement `update` function:** For Type 1 queries (flip range), use the segment tree's update function to flip the bits within the given range. This involves traversing the tree, applying lazy propagation where needed, and updating the tree nodes.
5
Step 5: **Process Type 2 queries:** For Type 2 queries (update nums2), multiply the sum of 1s in `nums1` (obtained from the root of the segment tree) by `p` and add it to the running sum of `nums2`.
6
Step 6: **Process Type 3 queries:** For Type 3 queries (sum nums2), simply append the current sum of `nums2` to the result array.
7
Step 7: **Return results:** Finally, return the array containing the results of all Type 3 queries.

💡

Pro Tip for Students

Try tracing through each step with a small example on paper before coding. Understanding "why" each step happens is more important than memorizing the code!

⚡

Complexity Analysis

⏱️

Time Complexity

O(q*log(n))

💾

Space Complexity

O(n)

📦 Linear space - Memory grows with input size.

📝

Why?

The segment tree has a height of log(n), so traversing it during update or query operations takes O(log n) time. The space complexity is O(n) because the segment tree and lazy array both have sizes proportional to n.

📖New to Big O? Click to learn more

Big O Notation - Quick Guide

O(1)Best - Instant

O(log n)Excellent - Binary Search

O(n)Good - One loop

O(n log n)OK - Sorting

O(n²)Slow - Nested loops

O(2ⁿ)Very Slow - Recursion

Alternative Approaches

Brute Force

Iterate through the `nums1` array directly for each Type 1 query and flip the bits. For Type 2 queries, iterate through `nums1` and `nums2` and update `nums2`. Calculate the sum of `nums2` for Type 3 queries.

Time: O(q*n), where q is the number of queries and n is the size of the input arrays.Space: O(1)

Trade-offs: The brute force approach is simpler to implement but has a much higher time complexity, making it unsuitable for large input sizes. Use brute force only if n and q are very small.

✨

Key Insights

1Insight 1: Using a naive approach of iterating through the range for each update (type 1 query) would lead to a time complexity of O(n*q), where n is the length of the arrays and q is the number of queries, potentially exceeding the time limit for larger inputs. Therefore, a Segment Tree with Lazy Propagation is crucial.
2Insight 2: The Segment Tree allows for efficient range updates (flips in this case) and range queries (sum of 1s). Lazy propagation ensures that updates are applied to the tree nodes only when necessary, optimizing the process.
3Insight 3: The sum of nums2 can be maintained throughout the operations, updated each time a type 2 query happens.

📚

Key Terms Explained

For Students

Array

A collection of items stored at contiguous memory locations. Like a row of boxes.

Edge Cases

Edge case 1: Empty ranges in Type 1 queries (l > r). The code implicitly handles this case because segment tree update ignores the call if the start > r or end < l.
Edge case 2: Queries with large values of 'p' in Type 2, ensure to use long data type to avoid integer overflow during the calculation of nums2 sum.

⚠️

Common Mistakes to Avoid

✗Common mistake 1: Not using Lazy Propagation with the Segment Tree. Without it, the time complexity for Type 1 queries becomes O(n) for each query, leading to TLE.

✗Common mistake 2: Integer overflow when calculating the sum of `nums2` or when updating `nums2` with `p * nums1[i]`. Use `long` data type to prevent it.

✗Common mistake 3: Forgetting to push the lazy updates down the tree before performing update or query operations.

✅

Learning Tip

Every mistake is a learning opportunity! When you make an error, understand why it happened before moving on.

🌍

Where is This Used in Real Life?

💼Real-world scenario 1: Managing large datasets where range updates and aggregations are frequently required, such as financial data analysis or sensor data processing. For example, imagine a system monitoring stock prices, where you need to update a range of historical prices and then query the overall change.

🏦Real-world scenario 2: Image processing, where operations like inverting colors within a rectangular region or adjusting brightness levels are performed frequently. The range updates could correspond to changing pixel values within a given area.

💡Understanding real-world applications helps you see the "why" behind algorithms. It's not just about passing interviews—these concepts power the apps and websites you use every day!

Author's Notes

💡 Key Insight

When dealing with arrays, always consider if sorting would simplify the problem. Many array problems have O(n) solutions using hash maps or two-pointer technique.

🎯 Interview Tip

It's okay to not finish hard problems in interviews. Show your problem-solving process.

⚠️ Watch Out

Off-by-one errors are the most common mistake. Double-check your loop bounds and array indices.

Written by Saswata Mondal

Software Engineer • Algorithms Enthusiast • Interview Coach

👨‍💻

About the Author

👨‍💻

Saswata MondalAuthor400+ LC Problems

B.Tech CSE, IEM Kolkata | LeetCode Rating: 1672 (Top 15%)

Certified: IBM Java Developer, NPTEL Programming in Java

Last updated: July 20, 2024LeetCode Profile GitHub

This solution has been crafted based on my experience solving 400+ LeetCode problems. Each explanation is designed to help you understand the underlying concepts, not just memorize the code.

Community Tips & Insights

In a real interview, always clarify constraints first. Ask about input size and value ranges.

@TechLeadJul 14

Don't forget to handle the edge case when the input is empty. Many candidates miss this in interviews.

@DevExpert92Jul 7

The key insight here is recognizing the pattern. Once you see it, the solution becomes straightforward.

@AlgoMasterJun 30

❓

Frequently Asked Questions

When should I use in-place modification vs creating a new array?

Use in-place modification when space complexity matters (O(1) space requirement) and the original array can be modified. Create a new array when you need to preserve the original data or when the problem involves significant restructuring that would complicate in-place logic.

Is it okay to not fully solve hard problems in interviews?

Yes! Interviewers care more about your problem-solving process than the final answer. Demonstrate: clear communication, systematic approach, ability to recognize when to pivot strategies, and graceful handling of hints. Partial progress with good reasoning is valued.

What is the importance of time and space complexity analysis?

Complexity analysis is crucial because: 1) Interviewers always ask about it, 2) It helps you choose between approaches, 3) It demonstrates CS fundamentals. Always state both time and space complexity, and be prepared to explain how you derived them.

What is the optimal approach for "Handling Sum Queries After Update"?

For this hard-level Array problem, the key is to identify the core pattern. Analyze the input constraints to determine acceptable time complexity. Consider whether segment tree techniques can optimize the solution. Always handle edge cases and validate your approach with examples before coding.

🤖 Need help with this problem?